Question: Simplify; express your answer in exponential form. Assume $r\neq 0, p\neq 0$. $\dfrac{{(r^{-4})^{-3}}}{{(r^{-3}p^{-1})^{3}}}$
Answer: To start, try working on the numerator and the denominator independently. In the numerator, we have ${r^{-4}}$ to the exponent ${-3}$ . Now ${-4 \times -3 = 12}$ , so ${(r^{-4})^{-3} = r^{12}}$ In the denominator, we can use the distributive property of exponents. ${(r^{-3}p^{-1})^{3} = (r^{-3})^{3}(p^{-1})^{3}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(r^{-4})^{-3}}}{{(r^{-3}p^{-1})^{3}}} = \dfrac{{r^{12}}}{{r^{-9}p^{-3}}}$ Break up the equation by variable and simplify. $\dfrac{{r^{12}}}{{r^{-9}p^{-3}}} = \dfrac{{r^{12}}}{{r^{-9}}} \cdot \dfrac{{1}}{{p^{-3}}} = r^{{12} - {(-9)}} \cdot p^{- {(-3)}} = r^{21}p^{3}$.